The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is :

  • Option 1)

    8\Omega

  • Option 2)

    16\Omega

  • Option 3)

    1\Omega

  • Option 4)

    6\Omega

 

Answers (1)

 

Heat developed in a resistor -

When a steady current flows through a resistance R for time t , the loss in electric potential energy appears as increased thermal energy(Heat H) of resistor and H=i^2Rt

The power devoleped = \frac{energy}{time}=i^2R=iR=\frac{V^2}{R}      (from Ohm's law)

- wherein

Unit of heat is joule (J)

Unit of power is watt (W)

 

 

P=\frac{v^{2}}{Req}

\Rightarrow Req=\frac{v^{2}}{P}=\frac{16^{2}}{4}=64\Omega

\Rightarrow 64=2R+R+4R+R

\Rightarrow 64=8R

\Rightarrow 8=R

\Rightarrow R=8\Omega


Option 1)

8\Omega

Option 2)

16\Omega

Option 3)

1\Omega

Option 4)

6\Omega

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